Yes, the performance would be worse. Would you ever notice it? I'd seriously doubt it. You mentioned running it in a cron job; you'd never see the effects of the lowered performance. The lack of added complexity might be worth it--less code (in this case) means easier to maintain. If you're worried about performance, you could always benchmark it--but I'd doubt it would make enough of a difference to worry about.
On Nov 26, 9:11 am, "Alex Jillard" <[EMAIL PROTECTED]> wrote: > Wouldn't that give me worse performance though? Filtering the two lists > verses making 300+ save()'s in a try/except, 90% of which won't be committed > to the database? > > On Wed, Nov 26, 2008 at 9:00 AM, Jeff FW <[EMAIL PROTECTED]> wrote: > > > If you already have a unique key on the project's name field, then > > you're good to go--no duplicates will ever get inserted. No need to > > do any filtering ahead of time--just put each save() in a try/except > > block that catches the error you get and does nothing. > > > -Jeff > > > On Nov 25, 4:53 pm, "Alex Jillard" <[EMAIL PROTECTED]> wrote: > > > Thanks for the replies Jeff and Rajesh. I'll look int both of those > > options > > > and see what I can come up with. My model is set to have the name of the > > > project to be unique, so it was throwing an error if a duplicate was > > added. > > > > Jeff, unfortunatly the initial list of CVS projects is just a string that > > > I'm splitting, so I can't do it as I build that list. > > > > Thanks again > > > > On Tue, Nov 25, 2008 at 4:38 PM, Rajesh Dhawan <[EMAIL PROTECTED] > > >wrote: > > > > > Hi Alex, > > > > > > Here is the code that I use. Output_list is from CVS and > > > > current_projects > > > > > is from the database. Is there a faster way to do this, or a built > > in > > > > > method in Python? This code works fine now, but I can see it getting > > > > slow. > > > > > > for b in output_list: > > > > > found = False > > > > > for i in current_projects: > > > > > if i.name == b: > > > > > found = True > > > > > if not found: > > > > > new_projects.append(b) > > > > > Try converting the two project name lists into Python sets and take a > > > > difference: > > > > > full_list = set(output_list) > > > > db_list = set([i.name for i in current_projects]) > > > > new_projects = full_list - db_list > > > > > You should also add a unique=True attribute to your project.name > > > > field. > > > > > -Rajesh D --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
